Electric potential on an infinite line of charge

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SUMMARY

The discussion focuses on calculating the electric potential at a radial distance of r = a/2 for an infinite line charge with a linear charge density of λ = +5 µC/m, situated within a thick conducting cylindrical shell. The potential is set to zero at the outer surface of the shell (V(b) = 0). The correct approach involves recognizing that the reference point for potential should be at r = a instead of r = b, leading to the correct application of the formula Vp - Vref = -∫E•dl = 2kλ*ln(Rref/R). The calculated potential at r = a/2 is approximately 88,176.55 V.

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  • Knowledge of integration techniques in physics
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Bryon
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Homework Statement


An infinite, uniform line charge with linear charge density λ = +5 µC/m is placed along the symmetry axis (z-axis) of an infinite, thick conducting cylindrical shell of inner radius a = 3 cm and outer radius b = 4 cm. The cylindrical shell has zero net charge.

The electrical potential is chosen to be zero at the outer surface of the cylindrical shell, V(b) = 0. (In this problem, it is not possible to chose the potential to be zero at infinity because the charge distribution extends to infinity.)

(c) Calculate the potential at a radial distance r = a/2.

?


Homework Equations



Vp-Vref = -∫E•dl = 2kλ*ln(Rref/R)

The Attempt at a Solution



Since the point b = 0.4 is the reference point I substituted that into the equation. Here is what I did:

R = 0.03/2 = 0.015

V = 2*(8.99x10^9)*(5x10^-6)*ln(0.04/0.015) = 88176.54985V

What did I miss?
 
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For a < r < b, E(r) = 0 .

∴ \int_{r=b}^{r}\vec{E}\cdot \vec{d\ell}=\int_{r=b}^{r=a}\vec{0}\cdot\vec{d\ell}+\int_{r=a}^{r}\vec{E}(r)\cdot\vec{d\ell}

The non-zero part of the integral begins at r=a.
 
I understand where the potential is zero but I cannot figure out the potential at r=a/2. Did I calculate it wrong?
 
Bryon said:
I understand where the potential is zero but I cannot figure out the potential at r=a/2. Did I calculate it wrong?
Yes, it's wrong.

I showed where your "formula": Vp-Vref = -∫E•dl = 2kλ*ln(Rref/R) comes from.

So, it's as if Rref = a, not b. Technically it's because V(b) = V(a).
 

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